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A ball is projected from a smooth horizontal surface with speed v at angle theta above the horizontal. The coefficient of restitution between the ball and the surface is e. Measuring along the horizontal, find the distance from the point of projection to the point where the ball strikes the surface for the SECOND time.

1. v² sin(2theta)(1 + e²) / g
2. v² sin(2theta)(1 + e⁴) / g
3. v² sin(2theta)(1 + e³) / g
4. v² sin(2theta)(1 + e) / g

Correct answer: v² sin(2theta)(1 + e) / g

Solution

The first range is R1 = v² sin(2theta)/g; after the bounce the time of flight scales by e (vertical speed becomes e*vsin theta) while horizontal speed is unchanged, giving R2 = e*R1, so total distance = R1(1 + e) = v² sin(2theta)(1 + e)/g.

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